Show that every eigenvalue to a unitary operator has absolute value 1

Question

Let $A$ be a unitary operator on hilbert space $H$, i.e. $$(Au|Av) = (u|v)$$ for all $u,v \in D_A.$.

I'm asked to show that all eigenvalues to this unitary operator has absolute value 1.

My attempt:

Let u,v be eigenvectors to $A$ and $\lambda_1, \lambda_2$ their eigenvalues.

$$(Au|Av) = (\lambda_1u|\lambda_2v) = \lambda_1\lambda_2 (u|v) = (u|v) \implies \lambda_1\lambda_2=1.$$

For the statement to be true, $\lambda_1$ must equal $\lambda_2$ but I can't see why this is true. What am I missing?

Answer 1

Let $\lambda$ and eigenvalue of $A$. With $u$ his eigenvector.

Then $(u,u) = (Au,Au) = (\lambda u, \lambda u) = \lambda\overline{\lambda}(u,u) = |\lambda|^{2}(u,u)$

Thus $|\lambda|^{2} = 1 \implies |\lambda| = 1$